network and subnets

[sexyali]

i'm trying to do the following problem,

my query is to work out the network address do i convert the host address to binary along with the supernet and add them together to get the answer

and then add the network address to subnet mask to get the subnet address/
thanks in advance

[adam999]

http://en.wikipedia.org/wiki/Subnetwork

[sexyali]

i've had a go on the first example,

here's wot i have worked out so far

network address = Host_ad + Subnet mask

10100100.00001110.01010011.00010011 (host_address)
11111111.11111111.11111111.00000000 (subnet_mask)
-----------------------------------------------
10100100.00001110.01010011.00000000 (network address) = 164.14.83.0

Subnet Address = Net_ad + supernet

10100100.00001110.01010011.00000000 (network Address)
11111111.11111111.00000000.00000000 (Supernet)
-----------------------------------------------
10100100.00001110.00000000.00000000 (subnet address) = 164.14.0.0


have i dont this correctly?

[Teknyk]

A supernet is just a bunch of subnets in a row.
the network address for 164.14.83.19 using that supernet netmask is 164.14.0.0
(/16 = 255.255.0.0)

The subnet you would find using the 255.255.255.0 and that is 164.14.83.0

assuming subnet zero is allowed and they are all using that same subnetmask, out of that supernet you would have the following subnets;
164.14.0.0
164.14.1.0
164.14.2.0
etc up to..
164.14.255.0

if you want to get good at subnetting you will need to memorize these;
10000000 = 128
11000000 = 192
11100000 = 224
11110000 = 240
11111000 = 248
11111100 = 252
11111110 = 254
11111111 = 255

your next one is;
192.128.252.3 with a /20

Now all those 1's and 0's give me a headache (and it take too dang long) so here is what I do.

I know there are 8 bits in each octet and a /20 = 20 bits.
This tells me the first 2 octets are 255 (8 each equaling 16) and leaves me with 4 remaining bits. So far I have 255.255.
the chart above shows 4 bits = 240
now I have a netmask of 255.255.240.0 (you know the last part is zero because you ran out of bits) for 192.128.252.3 according to your supernet.

now here is where a real shortcut comes in handy.
in that netmask the only "interesting octet" is the third one with 240 in it. So I now want the third octet out of the IP address which is 252

subtract 240 from 256 and that leaves 16
how may times can you put 16 into 252 without going over?
15 times... and 15 x 16 = 240 (do not confuse this with your netmask though!)

so 192.128.252.3 with a /20 is:
192.128.240.0 network
255.255.240.0 netmask

the subnet of 192.128.252.3 with 255.255.255.0 is again easy, drop the last number from the IP and make it 0 so... 192.128.252.0 with subnet mask 255.255.255.0

again assuming subnet zero is allowed and the same subnet masks are used out of that supernet you would get the following subnets
192.128.240.0
192.128.241.0
192.128.242.0
etc up to..
192.128.255.0

I hope I didn't just do your homework but try it step by step with your next one.
If you get real confused you can msg me.